Acceleration of stacked blocks on an inclined plane

[Ark236]

Homework Statement

Hi everyone,

The problem consists of two stacked blocks, blocks A and B, that lie on an inclined plane at an angle theta. Block B is sliding on a rough surface whose static friction coefficient is muk. The block A is at rest over the block B, and between the surfaced of the blocks there is a static friction coefficient μs.

Relevant Equations

I want to determine the acceleration of each body (is the same). To do that, I determine the equation of motion for each body:

For the body B

sum fx^B: F_{r,B} - f_{r,A} - (m_A+m_B) g sin(theta) = - m_B a

sum fy^B: N_B - (m_A+m_B) g cos(theta) = 0

For the body A

sum fx^A: f_{r,A} - m_Ag sin(theta) = - m_A a

sum fy^A: N_A -m_A g cos(theta) = 0

In order to calculate the acceleration, I sum the equations sum fx^B and sum fx^A.

F_{r,B} - m_Ag sin(theta) - (m_A+m_B) g sin(theta) = - (m_A+m_B) a

then

a = -[F_{r,B} - (2 m_A+m_B) g sin(theta)]/(m_A+m_B)
=-[muk (m_A+m_B) g cos(theta) - (2 m_A+m_B) g sin(theta)]/(m_A+m_B)

but the correct answer is a = -[muk (m_A+m_B) g cos(theta) - (m_A+m_B) g sin(theta)]/(m_A+m_B). I don't know what my mistake is.

thanks

 

[TSny]

Does the gravitational force $m_A \vec g$ act on body B?

How many individual normal forces act on body B? Describe them. How would you modify your diagram to show these individual forces?

 

[kuruman]

In addition @TSny's remarks, since you know that the two masses accelerate as one, why not draw a combined FBD of the two-mass system? It will make your calculations simpler because you only have to consider the external forces of gravity plus the friction and normal force from the incline. It's as if the two masses were glued together.

 

[Ark236]

Hi,

Regarding TSny's answer. For me is correct, the force of body A that acts on body B is m_A g. In the Body B there is one normal, N_b, that is the reaction of the surface between the body B and the plane.

Regarding Kuruman's answer, if I do it that way I immediately get the correct result. But I don't understand why if I do it the original way, I don't get the correct result.

 

[TSny]

Regarding TSny's answer. For me is correct, the force of body A that acts on body B is m_A g.

Suppose a block sits on the floor of an elevator that is accelerating downward. The floor exerts a normal force on the block while the block exerts a normal force on the floor. Are these forces equal in magnitude? Why? Does the magnitude of the normal force that the block exerts on the floor equal the weight of the block? Consider the extreme case where the elevator is in free fall.

In the Body B there is one normal, N_b, that is the reaction of the surface between the body B and the plane

What about the contact of the surface of body A with the surface of body B?

 

[Ark236]

Thanks, I think it's a little clearer now. The problem is that the force that body A exerts on body B is not m_A g since the system is accelerating. However, for me is still not easy to see to assume that, because similar to the elevator problem the person is exercising the real weight over the balance , i.e m g, and the reaction of the balance is modify by the aceleration of the system.

 

[kuruman]

Regarding Kuruman's answer, if I do it that way I immediately get the correct result. But I don't understand why if I do it the original way, I don't get the correct result.

You say you got the correct result when you considered the two masses together. Fair enough. Then to verify that your two FBDs are correct when you consider the masses separately, add them. The action-reaction pairs should cancel out and all that is left must be the FBD for the combined masses that you already have. To show you what I mean see the figure below modified from yours.

I see immediately the issue addressed by @TSny

Does the gravitational force $m_A\vec g$ act on body B?

You know that the combined weight of the two masses should be $(m_A+m_B)\vec g.$ Is that what you get when you add the weights in the two FBDs?

I also see that the green arrow pointing to the left in the A diagram does not belong. Acceleration is not a force and does not belong in a FBD unless you are considering a non-inertial frame.

 

[Tomy World]

 

[Tomy World]

 

[Tomy World]

Homework Statement: Hi everyone,

The problem consists of two stacked blocks, blocks A and B, that lie on an inclined plane at an angle theta. Block B is sliding on a rough surface whose static friction coefficient is muk. The block A is at rest over the block B, and between the surfaced of the blocks there is a static friction coefficient μs.
Relevant Equations: I want to determine the acceleration of each body (is the same). To do that, I determine the equation of motion for each body:

For the body B

sum fx^B: F_{r,B} - f_{r,A} - (m_A+m_B) g sin(theta) = - m_B a

sum fy^B: N_B - (m_A+m_B) g cos(theta) = 0

For the body A

sum fx^A: f_{r,A} - m_Ag sin(theta) = - m_A a

sum fy^A: N_A -m_A g cos(theta) = 0

In order to calculate the acceleration, I sum the equations sum fx^B and sum fx^A.

F_{r,B} - m_Ag sin(theta) - (m_A+m_B) g sin(theta) = - (m_A+m_B) a

then

a = -[F_{r,B} - (2 m_A+m_B) g sin(theta)]/(m_A+m_B)
=-[muk (m_A+m_B) g cos(theta) - (2 m_A+m_B) g sin(theta)]/(m_A+m_B)

but the correct answer is a = -[muk (m_A+m_B) g cos(theta) - (m_A+m_B) g sin(theta)]/(m_A+m_B). I don't know what my mistake is.

thanks

Hi, everything you did was correct, except that m_A should be not included in this free body diagram analysis. Review this. That's causing error in your calculation.

 

[kuruman]

Hi, everything you did was correct, except that m_A should be not included in this free body diagram analysis.

That is not true. Missing from OP's diagram is the downward normal force exerted by the top on the bottom block. OP's diagram for the top block also includes the acceleration as a force which shouldn't be.

 

[Tomy World]

That is not true. Missing from OP's diagram is the downward normal force exerted by the top on the bottom block. OP's diagram for the top block also includes the acceleration as a force which shouldn't be.

Haha, fair comment! Clear sign convention and learning the right methodology are more important than getting the right answer.

 

[Lnewqban]

... However, for me is still not easy to see to assume that, because similar to the elevator problem the person is exercising the real weight over the balance , i.e m g, and the reaction of the balance is modify by the aceleration of the system.

If the balance is accelerating downwards, the person is unable to transfer “the real weight” that had been measured in repose state onto the balance.
That real weight is always acceleration dependent.

Stoped elevator ⇒ Person’s weight = mg
Ascending accelerating elevator ⇒ Person’s weight = mg + ma
Descending accelerating elevator ⇒ Person’s weight = mg - ma

The extreme of that last situation is a vertical acceleration (downwards elevator and balance) that reaches the value of g; the person is weightless in free fall (reason for which he/she feels that weird feeling in the belly).

Free-falling accelerating elevator ⇒ Person’s weight = mg - mg = 0

 

[nasu]

Thanks, I think it's a little clearer now. The problem is that the force that body A exerts on body B is not m_A g since the system is accelerating.

Even if the system is not accelerating, the force exerted by A on B is not the weight but the normal component of the contact force between the objetcs. It may happen that some normal force, in some situation, is equal to some weight. But this does not make weight and contact force the same force.

It may help if you think about force as always the result of the interaction between two bodies. The weight of A is due to the interaction of A with the Earth. The only forces due to this interaction is the weight of A and the opposite and equal force acting on the Earth. There is no other object on which the force due to interaction between A and Earth acts.
The interaction between A and B is described by what we call contact force. The tangential component is the friction and the normal component is the "normal force". As for any interaction, there are two equal and opposite forces, one acting on A and the other on B. None of these is due to gravitational interaction. For the normal component it more like an elastic-type force.
It is true that the value of the normal force depends on the weight of A, in this setup. But "depends on" does not mean "equals to".

There is no such think as a force being transfered in the sense that it acts on an object that is not one the interacting pair described by that force.

 

[Ark236]

Thank you very much for the help